Determine the equation of the quadratic function with x-intercepts -2 and 6, and minimum value -10

Question

please help me to place these numbers into an equation

Answer 1

f(x) = a(x - h)² + k

 

Vertex is (h,k)

 

You know the y-coordinate of the minimum, which is at the vertex.  The x-coordinate of the vertex must be midway between the zeros, so the vertex is at (2,-10).

 

h = 2

k = -10

 

f(x) = a(x - 2)² - 10

 

f(6) = 0 ⇒ 

 

0 = a(6 - 2)² - 10

 

a = 10/16 = 5/8

 

f(x) = (5/8)(x - 2)² - 10

 

Test the answer.

 

Answer 2

y = a(x+2)(x-6) has the desired given roots (x-intercepts).  minimum is at vertex.

use this fact, and the fact that (2,-10) is the vertex, to solve for a.

 

The x coordinate of the vertex is the average of the two roots.  Do you get a=5/8?

 

 

Answer 3

Okay, this one has a few steps but they aren't all that bad. First, the X-intercepts of -2 and 6 mean that when you factor the polynomial, 

X = -2      ,  X = 6          ...and before that step, the factors would have read:

X + 2 = 0,  X - 6 = 0

Put them together and you get:

Y = a(X + 2)(X - 6)

Foil together and get:

Y = a(X² - 4X -12)

Then distribute:

Y = aX² - 4aX - 12a

Okay, now remember that the formula for the vertex, which is what the lowest or highest point must be, isa

X = -B / 2A   = -(-4a) / 2(a) = 4a / 2a = 2

This means we have the point (2, 10)

Let's plug it into the equation we have so far:

10 = a(2² - 4*2 - 12)

10 = a(4 - 8 - 12)

10 = a(-16)

/-16   /-16

-5/8 = a

So the final equation is:

Y = (-5/8)(X² - 4X - 12)

Distribute and you are done!